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I am stuck on what seems like a completely intuitive proof.

A is a subset of X. ( + for disjoint union)

I need to show, first, that

(i) Closure of S = interior of s + boundary of S

Then I am asked to show that

(ii) X = interior of S + boundary of S + interior of the complement of S.

It's very understandable when you draw X and its subsets neatly on a piece of paper. But I don't know how to start this proof. Does it help that I have already shown earlier that

boundary of S = closure of S ∩ closure of the complement of S ?

Juniper Leaf
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1 Answers1

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For (i), you know that that both the interior and boundary of $S$ are contained in its closure. Now, if $x$ is an interior point of $S$, then a small neighbourhood of $x$ is contained in the interior of $S$ as well. If $y$ is a boundary point of $S$, then every small neighbourhood of $y$ has some point outside of $S$. Can you see why these two conditions are mutually exclusive?

For (ii), you know $X=A\cup A^c$ for any subset $A$. Now let $A=\bar S$. Then $$ X=\bar S\sqcup(\bar S)^c=\text{int}(S)\sqcup\partial S\sqcup(\bar S)^c. $$ Now you need to prove that $(\bar S)^c=\text{int}(S^c)$. Try using De Morgan's laws on $\bar S$.

Ian Coley
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  • (ii) is absolutely clear now. I'm not completely sure I follow the last bits of (i). If e-neighbourhood of Y which is a boundary point of S intersects some S', then the boundary of S is not open in S. Therefore it is closed in S, so together with the interior forms a closure of S? – Juniper Leaf Feb 13 '14 at 16:45